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### Answers (1)

Answer : $\frac{1}{5} x^{4}+\frac{c}{x}$

Give : $x \frac{d y}{d x}+y=x^{4}$

Hint : Using $\int x^{n}dx$

Explanation: $x \frac{d y}{d x}+y=x^{4}$

Divide by x

\begin{aligned} &\frac{d y}{d x}+\frac{y}{x}=\frac{x^{4}}{x} \\ &\frac{d y}{d x}+\left(\frac{1}{x}\right) y=x^{3} \end{aligned}

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{x} \text { and } Q=x^{3} \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{1}{x} d x} \\ &=e^{\log x} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d c=\log |x|+C\right] \\ &=x \end{aligned}

Hence, the solution of differential equation is

\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y x=\int x^{3} x d x+C \\ &=y x=\int x^{4} d x+C \\ &=y x=\frac{x^{5}}{5}+C \end{aligned}

Divided by x, we get

$=y=\frac{x^{4}}{5}+\frac{C}{x}$

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